Normal Matrices Suppose $A \in \mathbb{C}^{n \times n}$ and $B \in \mathbb{C}^{m \times m}$ are normal, and 
$M \in \mathbb{C}^{n \times m}$. Prove that $AM = MB$ if and only if $A^*M = MB^*$.
This is what I have so far. Since $A$ and $B$ are normal matrices, we know that $A^* = UA$ for some unitary $U$. Similarly, $B^* = BV$ for some unitary $V$. Then, assuming that $A^*M = MB^*$, we obtain that $UAM = MBV$.
 A: Because they are normal, $A=UDU^*$ and $B=VEV^*$, with $U,V$ unitary and $D,E$ diagonal and invertible. We have $AM=MB$, which now we can write
$$
UDU^*M=MVEV^*.
$$
Multiplying by $U^*$ on the left and by $V$ on the right, 
$$
DU^*MV=U^*MVE.
$$
Let $N=U^*MV$. So we have $$\tag{*}DN=NE.$$ Entry-wise, because $D$ and $E$ are diagonal,
$$
(DN)_{kj}=\sum_sD_{ks}N_{sj}=D_{kk}N_{kj},
$$
and
$$
(NE)_{kj}=\sum_s N_{ks}E_{sj}=E_{jj}N_{kj}.
$$
We have, thus, for all $k,j$. 
$$
D_{kk}N_{kj}=E_{jj}N_{kj}.
$$
So, when $N_{kj}\ne0$, $D_{kk}=E_{jj}$. Now
$$
(D^*N)_{kj}=\overline{D_{kk}}\,N_{kj}=N_{kj}\overline{E_{jj}}=(NE^*)_{kj}
$$
(since, when $N_{kj}\ne0$ we have $D_{kk}=E_{jj}$ and so their conjugates are equal; when $N_{kj}=0$ the equality holds anyway). We have shown that $D^*N=NE^*$. From the definition of $N$, 
$$
D^*U^*MV=U^*MVE^*.
$$
Then
$$
A^*M=UD^*U^*M=U(D^*N)V^*=U(NE^*)V^*=UU^*MVE^*V^*=MVE^*V^*=MB^*.
$$

The infinite-dimensional version of this result requires a much more technical proof, and it is known as the Fuglede-Putnam Theorem. 
